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An ideal gas molecule is present at 27^@...

An ideal gas molecule is present at `27^@C` . By how many degree centigrade its temperature should be raised so that its `U_(rms) , U_(mp)` and `U_(av)` all may double

A

`900^@C `

B

`108^@C`

C

`927^@C `

D

`81^@ C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine how much we need to raise the temperature of an ideal gas so that its root mean square velocity (Vrms), most probable velocity (Vmp), and average velocity (Vav) all double. ### Step-by-Step Solution: 1. **Convert Initial Temperature to Kelvin**: The initial temperature is given as \(27^\circ C\). To convert this to Kelvin, we use the formula: \[ T(K) = T(C) + 273 \] Substituting the value: \[ T = 27 + 273 = 300 \, K \] **Hint**: Remember to always convert Celsius to Kelvin when dealing with gas laws. 2. **Understand the Relationship of Velocities with Temperature**: The formulas for the different velocities are: - \(V_{rms} = \sqrt{\frac{3RT}{M}}\) - \(V_{mp} = \sqrt{\frac{2RT}{M}}\) - \(V_{av} = \sqrt{\frac{RT}{\pi M}}\) All these velocities are directly proportional to the square root of the temperature (\(T\)). 3. **Determine the Temperature Required to Double the Velocities**: If we want to double the velocities, we set up the following relationship: \[ V' = 2V \implies \sqrt{T'} = 2\sqrt{T} \] Squaring both sides gives: \[ T' = 4T \] Substituting \(T = 300 \, K\): \[ T' = 4 \times 300 = 1200 \, K \] **Hint**: When doubling a quantity that is proportional to the square root, remember to square the factor by which you want to increase it. 4. **Convert the New Temperature Back to Celsius**: Now we need to convert \(1200 \, K\) back to Celsius: \[ T(C) = T(K) - 273 \] Substituting the value: \[ T = 1200 - 273 = 927 \, ^\circ C \] **Hint**: Always convert back to Celsius after performing calculations in Kelvin, especially when the final answer requires Celsius. 5. **Determine the Increase in Temperature**: The increase in temperature from the original \(27^\circ C\) to \(927^\circ C\) is: \[ \Delta T = 927 - 27 = 900 \, ^\circ C \] ### Final Answer: The temperature should be raised by \(900^\circ C\) to double the root mean square velocity, most probable velocity, and average velocity of the ideal gas.
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